(a) Differentiate \(y = x^{\frac{1}{2}} + k^2 x^{-\frac{1}{2}}\) to find \(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}k^2 x^{-\frac{3}{2}}\).

Set \(\frac{dy}{dx} = 0\) to find critical points: \(\frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2}k^2 x^{-\frac{3}{2}}\).

Simplify to get \(x = k^2\). Substitute back into the original equation to find \(y = 2k\).

Thus, the minimum point is \((k^2, 2k)\).

(b) At \(x = 4k^2\), find \(\frac{dy}{dx} = \frac{1}{4k} - \frac{1}{16k} = \frac{3}{16k}\).

The y-coordinate at \(x = 4k^2\) is \(y = 2k + \frac{k}{2} = \frac{5k}{2}\).

The equation of the tangent is \(y - \frac{5k}{2} = \frac{3}{16k}(x - 4k^2)\).

When \(x = 0\), \(y = \frac{7k}{4}\).

(c) Integrate \(\int (x^{\frac{1}{2}} + k^2 x^{-\frac{1}{2}}) \, dx\) to get \(\frac{2}{3}x^{\frac{3}{2}} + 2k^2 x^{\frac{1}{2}}\).

Apply limits from \(x = \frac{9}{4}k^2\) to \(x = 4k^2\).

Calculate the definite integral: \(\left[ \frac{16k^3}{3} + 4k^3 \right] - \left[ \frac{9k^3}{4} + 3k^3 \right] = \frac{49k^3}{12}\).